Hi,

In the example

Maximum Fiber RTT = 3msec - (40usec + 10usec+2,7usec+4usec) = 243,3usec.

Hi. Jesse

We assure the calculation result is correct as this:

Maximum Fiber RTT = 3msec - (40usec + 10usec+2.7msec+4usec)

                              = 3,000usec - (40usec + 10usec + 2,700usec + 4usec)

                              = 3,000usec - 2,754usec

                              = 246 usec

Thanks for your question.

Hi,

It seems you use 10us/Km = 1s/10e5 Km = 10e5 Km/s as the light speed. But the more accurate light speed in fibers would be 2 x 10e5 Km/s. Is there a conservative estimation or something?

Hi. Liu

Let us answer your question as below:

10us/km is an abbreviated parameter to calculate an one-way "Maximum fiber distance" by multiplying a one-way transmission delay time per km(5usec/km) by 2. So, the above fomula could be re-expressed as "Maximum RTT / (2x5usec/km)".

Thus, the light speed over optic fiber can be calculated as "1/(one-way transmission delay time)= 1/(5usc/km) = 2x10E5km/s".

As you mentioned, 2x10E5km/s is the light speed over optic fiber cable.

I hope your question is asnwered by this comment.

Many questions from you will be welcomed. Thanks.

Hi Netmanias,

Delay components (Table 2) do not take into account (d):Active equipment processing and (j):Fiber latency (BBU to RRH) of Table 1. Why & if this is right, what is the point then of having (d) & (j) shown in Table 1?

Tnx,

Dimitris

Hi. Dimitris

Thank you for your question and comment on this article.

There was a mis-typing on the table as this: "c+k" should be "d+k".

We've corrected the mis-printed character directly now.

Sorry for making you confused by the error.

As you got a point, the fiber RTT can be calculated by subtracting a total of the delay components from 3msec. and then, you can estimate a maximum fiber distance (one-way) by dividing the calculated fiber RTT by 10us(=2x5us).

Thanks,

Netmanias

hello

according to :

Maximum Fiber distance (Km) = Maximum Fiber RTT / (10 µsec/Km)

in some article I studied that the distance is between 20 - 40 Km. can you please say how can I reach 40Km?

and what is the diffrent between PON 10 Gbps and PON 100 Gbps over this formula?

Hi,

Eventually I'm missing something but I've one basic question:

In the example:

Maximum Fiber distance (Km) = Maximum Fiber RTT / (10 µsec/Km) = 24,6 Km

Since one is considering "round trip time" and taking in account Table2' variables, than "24,6 Km" should correspond to the trip from RRH (cell site) to BBU (CO) and back again to RRH (cell site), shouldn´t it? The "Maximum fiber distance between RRH (cell site) and BBU (CO)" shouldn´t be half of 24,6 Km, i.e. 12,3 Km?

Thanks.

BR,

Hello carlos, 

according to "5 THINGS YOU SHOULD KNOW ABOUT FRONTHAUL" and "Cloud RAN for Mobile Networks - a Technology Overview" the distance is between 20-40Km

cheers.

Hi,

3ms is the delay for how many bytes data (is it for a radio frame etc.)? Did you consider transmission delay which depends on the bandwidth of fronthaul? Your response is much appreciated. 

Hi ,

What is about interoperability between BTS and transmission vendors with CPRI over Microwave case? 

Anyone has tried to integrate RRUS over E-band CPRI link with 6601 BTS? 

I see that some vendors like ALU allow to configure additional delay at BBU side to let CPRI synchronize over E-Band microwave link.

But I have got stuck with Ericsson as their BTS does not have an option.

Any advise ? 

Thanks,

Dmitry